Problem 2009

Let $\angle YXZ$ denote the directed angle between lines $XY$ and $XZ$, measured modulo $180^\circ$.

Since $OE = OF$, we have $\angle FAO = \angle OAE = \angle OAC$. Hence
$$\begin{aligned} \angle AOF &= 180^\circ - \angle FAO - \angle OFA \\ &= 180^\circ - \angle OAC - \angle OEC \\ &= 180^\circ - \angle OAC - \angle ACO \\ &= \angle COA, \end{aligned}$$
which yields that triangles $AOF$ and $AOC$ are congruent. Thus $AB = AC = AF$, and since $A$ is the circumcenter of triangle $BFC$, we have $\angle FAC = 2\angle FBC$.

Let line $ED$ meet $\Gamma$ again at $P$, then we have $\angle FPO = \angle OPE = \angle OPD$. Hence
$$\begin{aligned} \angle POF &= 180^\circ - \angle FPO - \angle OFP \\ &= 180^\circ - \angle OPD - (180^\circ - \angle PEO) \\ &= 180^\circ - \angle OPD - (180^\circ - \angle ODE) \\ &= 180^\circ - \angle OPD - \angle PDO \\ &= \angle DOP, \end{aligned}$$
which yields that triangles $POF$ and $POD$ are congruent, and it follows that $PD = PF$.

Now we have $\angle FPD = \angle FPE = \angle FAE = \angle FAC = 2\angle FBC = 2\angle FBD$. Then from this and $PD = PF$, it follows that the circumcenter of triangle $BDF$ is $P$, and lies on $\Gamma$.

Comment. If we don't consider the direction of the angle, to derive that $P$ is the circumcenter of triangle $BDF$ from $PD = PF$ and $\angle FPD = 2\angle FBD$, we need to show that $B$ and $P$ are on the same side with respect to line $DF$.

Another solution.

Let $a = \angle CBA = \angle ACB$, then we have $\angle EOD = 2\angle ECD = 2\angle ACB = 2a$. Since $\angle ODC = \angle OCD < \angle ACB = \angle ABC$, lines $AB$ and $OD$ are not parallel. Let these lines meet at $G$, then we have $\angle GAE + \angle EOG = (180^\circ - \angle CBA - \angle ACB) + \angle EOD = (180^\circ - 2a) + 2a = 180^\circ$, which yields that $G$ lies on $\Gamma$.

Note that $\angle ODE = 90^\circ - \frac{1}{2}\angle EOD = 90^\circ - a$. Let $P$ be the circumcenter of triangle $GBD$, then we have $\angle GDP = 90^\circ - \frac{1}{2}\angle DPG = 90^\circ - \angle DBA = 90^\circ - a = \angle ODE$, which yields that points $E, D, P$ are colinear. In addition, since $\angle PGO = \angle PGD = \angle GDP = \angle ODE = \angle DEO = \angle PEO$, $P$ lies on $\Gamma$.

Let $F'$ be symmetric to $D$ with respect to line $OP$. Since $OD = OF'$, $F'$ lies on $\omega$. Moreover, since $PB = PD = PF'$, $P$ is the circumcenter of $BDF'$. Now since $\angle OF'P = \angle PDO$ and $\angle PEO = \angle ODE$, it follows that $\angle OF'P + \angle PEO = \angle PDO + \angle ODE = 180^\circ$. Hence $F'$ lies on $\Gamma$. From this, $F'$ is the second intersection of $\Gamma$ and $\omega$, which yields that $F'$ coincides with $F$. Therefore the circumcenter of triangle $BDF$ is $P$ and lies on $\Gamma$.